Tuesday 26 November 2019

quantum field theory - Feynman $ivarepsilon$-prescription in path integral by adding an imaginary part to time


It is known that the well-definiteness of the path integral leads to the Feynman's $i\varepsilon$-prescription for the field propagator. I've found many ways of showing this in the literature, but it is precisely the way that I have learned in my QFT course (and which I have not found in literature) that I do not understand.


Context of the problem


Considering the case of a real scalar field for simplicity, one has that the following path integral (evaluated at asymptotic times)


\begin{equation} \lim_{T \rightarrow \infty}\int_{\phi(-T, \vec{x})}^{\phi(T,\vec{x})} \mathcal{D}\phi \ \text{exp} \left( i \int^T_T dt \int d³ x \ ( \mathcal{L}+J \phi) \right)\tag{1} \end{equation}


can be expressed as



\begin{equation} \lim_{T \rightarrow \infty} \sum_{m, n} e^{-i\left(E_n+E_m \right)T} <\phi, T|n, T>_J _J \tag{2} \end{equation}


where $|n>_J$ are eigenstates of the hamiltonian $H$ in the pressence of the source $J$. In order to make this oscillatory exponential converge (and properly define the path integral) one adds to $T$ a small imaginary part $T \rightarrow T(1-i\varepsilon)$. With this, one writes the vacuum persistence amplitude as \begin{equation} <0|0>_J = \frac{1}{N} \lim_{\varepsilon \rightarrow 0} \ \lim_{T \rightarrow \infty(1-i\varepsilon)} \int \mathcal{D} \phi \ \text{exp} \left( i \int_{-T}^T dt \int d³x (\mathcal{L}+J\phi)\ \right) \equiv \frac{1}{N} Z[J]\tag{3} \end{equation} where the constant $N$ is typically taken to be $N=Z[0]$.


My problem


In order to relate $Z[J]$ with the Feynman propagator $D_{F}(x-y)$, one typically writes the argument of the exponential in the Fourier space, then makes the change of variable $$\hat{\phi}(p)'=\hat{\phi}(p)+(p²-m²)^{-1} \hat{J}(p)\tag{4}$$ (which leaves $\mathcal{D}\phi'=\mathcal{D}\phi$) to get


\begin{equation} Z[J]=Z[0] \text{exp} \left(-\frac{i}{2} \int \frac{d⁴ p}{(2\pi)⁴} \hat{J}(p) \frac{1}{p²-m²} \hat{J}(-p)\right).\tag{5} \end{equation} Here I'm using the $(+,-,-,-)$ Minkowski sign convention.


Now here I've been told that the fact of replacing $T\rightarrow (1-i\varepsilon)T$ to define the path integral is equivalent in Fourier space as $p⁰ \rightarrow (1+i\varepsilon)p⁰$. With this, one gets the correct $i \varepsilon$ Feynman prescription for the propagator $D_F(x-y)$


\begin{equation} Z[J]=Z[0] \text{exp} \left(-\frac{i}{2} \int \frac{d⁴ p}{(2\pi)⁴} \hat{J}(p) \frac{1}{p²-m²+i\varepsilon} \hat{J}(-p) \right) = Z[0] \text{exp} \left(-\frac{1}{2} \int d⁴y \int d⁴x \ J(x) D_F(x-y) J(y) \right). \tag{6} \end{equation}


And this is the part that I don't get at all, I've tried but I don't see how the fact that $T\rightarrow (1-i\varepsilon)T$ leads in the previous approach to $p⁰ \rightarrow (1+i\varepsilon)p⁰$ and therefore to the Feynman prescription. I'm having nightmares with this, any help would be really appreciated.


NOTE: I use the following convention for the Fourier transform \begin{equation} \hat{\phi}(p)=\int d⁴ x \ \phi(x) e^{-ip \cdot x}\tag{7} \end{equation}


so that \begin{equation} \phi(x)= \int \frac{d⁴p}{(2 \pi)⁴} \ \hat{\phi}(p) e^{+ip \cdot x}.\tag{8} \end{equation}





No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...